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### Transcript of Introduction to Variational Methods and Finite Elements

• Introduction to variational methods and nite elements

1.2.3. Variational formulations of BVP:

Problem: Sove ax = b x =ba

Reformulate the problem:Consider E = 12ax

2 + bxFind x : E(x) = min

xE(x)

ax b

x

x

1. Rayleigh-Ritz Method:

Consider a dierential equation

Au = u = f(x) (1a)u(0) = u(1) = (1b)

} Functionalan dimensionvector

Consider the functional: E[u] =10

12(u

)2 + fu dx potential energy functional.

Claim: If u : E[u] = minu

E[u] and u satises (1b) then u solves (1).

Proof:

Let u = u + (arbitrary) (0) = 0 = (1) and C2 so we have a 1 parameter family offunctions which are perturbations of u. The function

Nu

E() = E[u + ] =1

0

12

(u

+

)2+ f(u + ) dx

E(0) =1

0

(u+ ) + f dx

=0

=

10

(u) + f dx

= u10

10

(u

f) dx = 0

22

• Since is arbitrary we could choose =(u

f)

u f = 0 Euler -Lagrange equation for E[u].

How is this useful? Let us assume that U(x) =N1k=1

kk(x)

basis functions

Then

E() =

10

12

[N1k=1

kk(x)

]2+ f(x)

Nk=1

kk(x) dx

For a min:

0 =E

j=

10

(N1k=1

kk

)j + f(x)j(x) dx

=N1k=1

k

10

kj dx+

10

f(x)j(x) dx, j = 1, . . . , N 1

A = b where Akj =1

0

kj dx bj =

10

f(x)j(x)dx.

Example : u = x2u(0) = 0 = u(1)

}u(x) =

x

12(1 x3)

Assume {j} ={1, x, x2, x3

}

U3(x) = c0 + c1x+ c2x2 + c3x3

U3(0) = c0 = 0 u3(1) = c1 + c2 + c3 = 0U3(x) = ax(1 x) + bx2(1 x)

= a{x x2}1

+ b {x2 x3}2

c2x+ c2x2 = c3x+ c3x3c2x(1 x)c2 = c1 c3

A11 =

10

(1 2x)2 dx =1

0

1 4x+ 4x2 dx = x 2x2 + 43x3

1

0

=13

A22 =

10

(2x 3x2)2dx =1

0

4x2 12x3 + 9x4 dx = 43x3 3x4 + 9

5x5

1

0

=20 45 + 27

15=

215

A12 = A21 =

10

(1 2x)(2x 3x2) dx = 16

b1 = +

10

x2(x x2) dx = 120

23

• b2 =

10

x2(x2 x3)dx = 130

13 1616

215

a

b

=

120

130

a = 1

15b =

16

U3(x) =x

30(1 x)(2 + 5x)

Exact:

uex(x) =x

12(1 x3)

Natural and Essential B-C

Notice that the basis functions {i} were required to satisfy the BC u(0) = 0 = u(1). B-C that haveto be forced onto the trial solution are called essential B-C typically these involve the solutionvalues and not the derivatives. In the case of derivative B-C it is possible to build the B-C into theenergy functional to be minimized.

Consider{

u(0) = u(1) = u = f(x)

Old energy functional Boundary Condition

E[u] =

10

12(u)2 + fu dx u(1)

E[u] =

10

uu + fu dx u(1)

= uu10

10

(u f) u dx u(1)

= {u(1) }u(1)1

0

(u f)u dx = 0

u = f and u(1) = but says nothing about u(0)!

Dont have to enforce BC at this endpoint.

Eg. 2:

u = x = f uexact = 2 + 7x2 x3

6u(0) = 2 u(1) = 3

essential natural

Let: U(x) = 2 + x(a1 + a2x) = 2 + a1x+ a2x2

24

• satises homogeneous version of essential BC

UN (x) = +N

i=1

aii(x) =N

i=0

aii(x) a0 = 1 0(x) = 2

E(a) =

10

12

(aii

)2+ f

(aii

)dx 3

(aii(1)

)

0 =E

aj=

10

(aii

)j + fjdx 3j(1)

0 =(i,

j

)ai + (f,j) 3j(1)

0 = Aa+ b

A11 =

10

dx = 1 A21 = A12 =

10

2x dx = 1 A22 =

10

4x2dx =43x3

1

0

=43

b1 =10

(x) x dx 3 = 313

b2 =10

(x)x2 dx 3 = 314[

1 11 4/3

] [a1a2

]=[

10/313/4

] a1 = 4312 a2 =

14

U2(x) = 2 +4312

x x2

4

x 0 0.2 0.4 0.6 0.8 1RR 2 2.707 3.393 4.060 4.707 5.333EX 2 2.699 3.389 4.064 4.715 5.333 Built in

2. General calculus of variations:

I[u] =

ba

F (x, y, y) dx y(a) = 0, y(b) = 0

0 = I =

ba

F

yy +

F

y(y) dx

= yF

y

b

a

b

a

[d

dx

(F

y

) F

y

]y dx

ddx

(F

y

) F

y= 0 Euler-Lagrange Eq.

Eg.1 (pu) + qu = fu(a) = u(b) =

I[u] =

ba

p(u)2 + qu2 2fu dx+ u

F = p(u)2 + qu2 2fuF

u= 2qu 2f

F

u= 2pu

x

(F

u

) F

u= 2

{(pu) qu + f} = 0

25

• Eg. 2: Sturm-Lioville Eigenvalue problems:

u(a) = 0 = u(b); +(pu) + qu + ru = 0 (1)

Multiply by v and integrateba+(pu)v + quv dx +

baruv dx = 0

= I[u] =

bap(u)2 qu2 dx

baru2 dx

=I1I2

=I1I2 I1I2

I22=

1I2

(I1 I2)

=

ba2puu 2quu ruu dx

baru2 dx

=2

ba

[(pu) + qu + ru

]u dx

baru2 dx

= 0

= 0 the function u0 which minimizes I[u] is an eigenfunction of (1) and = I[u0] is its eigenvalue.

Higher eigenvalues: n = min(u,k)=0

k=1,...,n1I[u]

constrained minimization problem.3. Method of weighted residuals

What do we do for nonlinear or dissipative problems for which potential energy functionals dontexist or cannot be found easily?

Consider a BVP

Lu = f in = (a, b) (1a)u(a) = ; u(b) + u(b) = (1b)

An approximate solution U wont in general satisfy (1a) and we associate with U the so-calledresidual

r(U) = LU fNote that r(u) = 0

exact A whole class of methods are obtained by considering various ways to minimize the residual insome sense, usually:

ba

r(U)i dx = 0

test or weight functions

where U =N

i=1

aii(x) basis function

26

• 1. Collocation: weight functions = (x xi) basis functions polynomials:

Eg:

u u = 0u(0) = 1, u(1) = 0

(A) Let i(x) =(xi+1 x)which satisfy the homogeneous B-C for the problem

U(x) = 1 x+ a1(x2 x)+ a2 (x3 x)

r(U) = {a12 + a26x} {a1 (2x 1) + a2

(3x2 1)}+ 1

= (3 2x) a1 a2(3x2 6x 1)+ 1

With x1 = 0 and x2 = 1 as collocation points we have(3 11 4

)(a1a2

)=( 11

) U1(x) = 1 x

(311

)(x2 x) 2

11(x3 x)

If we choose x1 and x2 are chosen to be the zeros of the 2nd degree Legendre polynomial

U2(x) = 1 x 14(x2 x)

(16

)(x3 x)

The exact solution is u(x) =(e ex)(e 1) .

x u(x) U1(x) U1 u U2 U2 u U3 U3 u1/4 0.834704 0.837750 3.05103 0.835938 1.23103 0.769965 6.47 1021/2 0.622459 0.636364 1.39 102 0.625000 2.54 103 0.506144 1.16 1013/4 0.349932 0.360795 0.351563 1.63 103 0.269965 8.00 102

(B) With dierent trial functions:

U(x) = 1 x+ a1 sinx+ a2 sin 3xr(U) = ( sinx+ cosx)a1 3(3 sin 3x+ cos 3x)c2U3(x) = 1 x+ 0.017189 sinx+ 0.011045 sin 3x

using zeros of 2nd degree Legendre polys.

Where to collocate?

What basis fucntions to use?

DeBoor, C. and Swartz, B. , Collocation at Gaussian points, SIAM J. Num. Anal. 10 (1973),582606.

27

• 2. Method of moments: weight functions i = xi.

Eg:

u + u + x = 0u(0) = 0 = u(1)

Let U(x) = a1x(1 x) + a2x2(1 x) + . . .r(U) = x+ a1(2 + x x2) + a2(2 6x+ x2 x3)1

0

r(U) 1 dx = 0 and1

0

r(U) x dx = 0

116 1112

1122

1920

a1

a2

=

12

13

U(x) = x(1 x)(122649

+110649

x

)

uex(x) =sinxsin 1

x

3. Galerkin method: Expand U(x) and w(x) in terms of the same basis functions.

u(x) =N

i=1

aii w(x) =N

i=1

bii

r

(N

i=1

aii

)(N

i=1

bii

)dx = 0

r

(N

i=1

aii

)j dx = 0

Since the bi are arbitrary.

Eg. u + u+ x = 0u(0) = 0 = u(1)

Let U(x) = a1x(1 x) + a2x2(1 x)r(U) = x+ a1(2 + x x2) + a2(2 6x+ x2 x3)1

0

r(u) 1(x)dx =1

0

{x+ a1

(2 + x x2)+ a2 (2 6x+ x2 x3)} x(1 x)x2(1 x) dx = 0

310

320

320

13105

a1

a2

=

112

120

a1 = 7369 a2 = 741

U(x) = x(1 x)(

71369

+741

x

)

ue(x) =sinxsin 1

x

28

• Relationship between Rayleigh Ritz and Galerkin:

Rayleigh Ritz u + u = f = x u(0) = 0 = u(1)

I[u] =

10

u2 u2 + 2fu dx

0 = I =

10

2uu 2uu + 2fu dx

=[uu

]10

10

(u + u f) u dx

0 =1

0

(u + u f)u dx

=

10

r(u)u dx

I(a) =

10

(aii

)(aij

)(

aii)(

ajj

+2f

ajj dx

0 =I

ak= 2

{ai(i,

k)

ai(i,k)

+(f,k)}f = x

Rayleigh-Ritz and Galerkin methods are identical for this problem. true in general for linearproblems but not for nonlinear problems.

From WR to the weak form:

WR1

0

(u + u+ x)v dx = 0

u(0) = 0 = u(1) v satisfy homogeneous Dirichlet i.e. v(0) = 0 = v(1)

Say we wanted to express u in terms of functions that are not twice dierentiable then we integrateby parts to throw as many derivatives as needed from u to v.

10

(u + u + x)v dx = uv

1

0

+

10

uv + uv + xv dx = 0.

Now if we let u =

aii and v =

bii we have since v is arbitrary

i

ai(i,j) +

ai(i,j) + (x,j) = 0

Aa = b

where[A]jj = (i,

j) (i,j); b = (x,j)

Identical to Rayleigh-Ritz eqs.

29

• What about a natural BC?

(S){

u + u+ x = 0u(0) = u(1) = .

10

(u + u+ x)v dx = 0 v H10

uv10

10

uv uv xv dx = 0

v(1)1

0

uv uv xv dx = 0

(W)

Find u H1 such that1

0

uv uv xv dx v(1) = 0 v H10

The Finite Element method

F.E. Model Problem

u + f = 0u(a) = G; u(b) = H

}(S)

N i(x) =

Ni(x) =W Residual

ba

(u + f

)v dx = 0 cant plug in

uh(x) =N

i=0uiNi(x)

N i (x) = (x xi1) 2(x xi) + (x xi+1)We must therefore relax the continuity required of the trial so-lution by going to a weak formulation of the problem.

uv

b

a

+

ba

uv + fv dx = 0

b

a

uv dx = Hv(b) +b

a

fv dx

a(u, v) = Hv(b) + (f, v) ()

(W) Given f L2 and constants G,H ndu {u H1, u(a) = g} = H1g such that for allv H10 = {v H1, v(a) = 0} we have thatbauv dx = Hv(b) +

bafv dx

Hk = {u : u(k) L2} Sobolov space

30

• Claim: If u is a solution of (S) then usatises (W ). Conversely, provided u is suciently dieren-tiable then if u satises (W ) it follows that u also satises (S).

(S) (W )( See *)(W ) (S)

ba

uv dx = Hv(b) +b

a

fv dx for all v H10

uv

b

a

b

a

uv dx = hv(b) +b

a

fv dx

ba

(u + f)v dx+ {H u(b)}v(b) = 0

v arbitrary u + f = 0 u(b) = H

natural BC

Using Finite Element Basis Functions: It is useful in the calculation of the integrals to trans-form each of the subintervals in turn to the same standard interval on which we construct cardinalbasis functions

x0 xi x x3

N2N1

N2()N1()

1 10

xi+1hi+1

N1() = 12(1 )

N2() = 12(1 + )

Na() = 12(1 + a)

a ={ 1 a = 1

1 a = 2

Transformation:

(x) = c1 + c2x (xi) = c1 + c2xi = 1 c2 = 2xi+1 xi =

2hi+1

(xi+1) = c1 + c2xi+1 = 1 c1 = (xi + xi+1)hi+1

(x) =2x (xi + xi+1)

hi+1

Similarly x() =hi+1 + (xi + xi+1)

2

31

• Notice we can express the transformation in terms of Na():

x() =2

a=1

Na()xa

dx

d=

2a=1

() =a2

)

=12

2a=1

axa =12(xi + xi+1) = hi+12

d

dx=

2hi+1

Galerkin approximation:

Let uh(x) = GN0(x) +N

i=1

Ni(x)ui

vh(x) =N

i=1

Ni(x)vi arbitrary

a(uh, vh

)= hvh(b) +

(f, vh

)

b

a

(N

i=1

uiNi(x)

) Nj=1

viNj(x)

dx = h vjNj(b) +

ba

fN

j=1

vjNj(x) dx

vj arbitrary N

i=1

ui(N i , N

j

)= hNj(xN ) + (Nj , f) j = 1, . . . , N

Ku = fstiness matrix

Kij =(N i , N

j

)=

ba

N iNj dx

fj = hNj(xN ) + (Nj , f)

i 2 i 1 i i + 1 i + 2

Notice :

(a) Kij = Kji Symmetric

(b) Kij is positive denite

1ijN

uiKijuj =ij

uia(Ni, Nj)uj = a(uh, uh

)=

ba

(u)2

dx 0 pos. semi def.

32

• not included

(uh)

= 0

= uh = const but since uh(0) = 0

= uh(x) =

uiNi(x) = 0

form a basis ui 0. uTKu 0 = 0 u = 0

(c) Kij = 0 if |i j| 2

Tridiagonal system - like nite dierences.

33

• The trick to doing the integrals is to calculate them element by element: so called assembly of thestiness matrix:

e 2 e 1 e e + 1Element stiness matrix

Kij =

ba

N iNj dx =

Ne=1

xexe1

N i(x)Nj(x) dx

=N

e=1

keab

keab =

xexe1

N a(x)Nb(x) dx

=

11

()(x)dNbd

(x) dxd

d

=2he

a2

b2

11

d =abhe

=(1)a+b

he

[keab] =1he

1 11 1

and keab = 0 if a, b {e 1, e}

Na() =12(1 + a)

N a() =a2

d

dx=

2xe xe1 =

2he

Assembly:

FIGURE assume he = h e i.e. uniform mesh

1h

2 11 2 1

. . .. . .

1 1

u1

...

un

=

F

To calculate the force for a general f(x):

fj =

ba

f(x)Nj(x) dx

We assume f(x) fh(x) =jfjNj(x)

The force vectors are also assembled

fj =e=1

xexe1

f(x)Na(x) dx

34

• fea =2

b=1

fb

xexe1

Nb(x)Na(x)dx

Simple Examples

01

N1N0

x1 x

(1) N = 1(a)

u = 0 f 0u(0) = G u(1) = G

}ue = G +Hx

k11 =11 1 = 1

f1 = H 1 +1

0

x 0 dx = H

u0 + u1 = f1 u1 = H + G uh=1(x) = GN0(x) + (G + H)N1(x) = G + Hx

(b)

f(x) = pu + p = 0 up = Ax2 2A+ p = 0

ue(x) = + x p2x2

u(0) = G = u(x) = px u(1) = p = H = (p +H)

ue(x) = G +Hx+ p(x x

2

2

)

f1 = H +

10

x p dx = H +p

2

k11u1 = H +p

2+ G

uh(x) = G + Hx+ px2

does not capture exact solution becauseue {set of piecewise linear functions on (0, 1)} .

however uh(0) = G uh(1) = G +H +p

2= ue(1) exact at nodes

(c)

f = qx

ue(x) = G + Hx+ qx

2

(1 x

2

3

)

f1 = H + q

10

x2dx = H +q

3

u1 = H +q

3+ G

35

• uh(x) = G + Hx+qx

3uh(1) = G + H +

q

3= ue(1) Galerkin exact at nodes

N1 N2

x2 = 1x1

h = 1/2

(2) N = 2 h = 12(a) f 0: u0 = G

1(1/2)

2 11 1

u1

u2

=

HN1(1) +10

N1 0 dx+ 2G

HN2(1) +10

N2 0 dx

4 22 2

u1

u2

=

2G

H

u1 = G +H

2u2 = G + H

uh(x) = GN0 +(G +

H

2

)N1(x) + (G +H)N2(x)

= G + Hx

(b) f = p:

4 22 2

u1

u2

=

2G + p10

N1 dx

H + p10

N2 dx

=

2G + p/2

H + p/4

u1

u2

=

3p/8 +G + H/2

p/2 + G +H

uh(x) = N0G +N1(x) (3p/8 + G +H/2) +N2(x) (P/2 + G +H)

= G + Hx+N1 3p8 +N2p

2

uh(1/2) = G +H

2+

3p8

= uexact(1/2)

uexact(x) = (G+ Hx+ p(x x2/2)

(c)

f = qx

uh = G + Hx+11q48

N1(x) +q

3N2(x)

uh(1/2) = G +H

2+

11q48

= uexact(1/2)

36

• Using a nite dierence approximation

u + f = 0 f = pu(0) = G u(1) = H

01 2 3

N = 2

n = 1 :U2 2U1 +G

h2= p 1

h[2U1 U2] = hp+ G

h

n = 2 :U3 2U2 + U1

h2= p U3 U1

2h= H U3 = U1 + 2hH

2U2 2U1 2hHh

= hp = U2 U1 = H + h2p

1h

2 11 1

U1

U2

= [ G/h + hp

H + h2p

]h = 1/2

Exactly the same equations as the FEM.

Some convergence results for the F.E. model problem:

We consideru + f = 0u(a) = G u,x(b) = G

}(1)

(A) The Greens function for (1)

Find G(x, x) : G,xx(x, x) + (x x) = 0G(a, x) = 0 = G,x(b, x)

G,x + H(x x) = c1G + x x = c1x+ c2

(i) G,x(b) +H(b x) = c10 + 1 = c1

(ii) G(a) + a x = a+ c20 + 0 = a+ c2

G(x, x) = (x a) x x

x

(x x)dx ={

0 x < x

1 x > x

= H(x x)x

H(x x)dx =

{0 x < x

(x x) x x

x a

a x b

G = x aG = (x a)

Note that G c0 is piecewise linear.

37

• Theorem 1: uh(x) the PWL Galerkin approximation is exact at the nodes; i.e., uh(xi) = uex(xi)

Notation: Let vh = {f span {N1, . . . , NN} where Ni H 0}.

Lemma 1: a(h, u uh) = 0 h vh

Proof:

(W) a(u, ) = (f, ) + H(b) H 0 ()u classical or strong solution(S) (W)

(G) a(uh, h) = (f, h) + Hh(b) () h vh =

{h =

ciNij ; Ni H 0

}vh H 0

() a(u, h) = (f, h) +Hh(b) ( )Subtract () from ( ):

a(u uh, h) = 0 h vha(h, u uh) = a(u uh, h) symmetry

Lemma 2: a(G(x x), ) = ((x x), ) H 0Proof:

Gxx + (x x) = 0G(a) = 0 = Gx(b)

0 =

b0

{Gxx + (x x)

} dx = Gx

ba+

ba

Gxx + dx H 0

a(G,) = ((x x), ) H 0

Lemma 3: u(x) uh(x) = a(G(x x) h, u uh) h vh.

Proof: u uh H 0u H1g and uh vh + {gN0} uh(a) = u(a) = g u uh H 0

H 0a(G(x, x

) h, u uh) = a(G, u uh) a(h, u uh)= ((x x), u uh) 0

L(2) L(1)= u(x) uh(x)

Proof of Theorem: u(xi) = uh(xi)

Let x = xi. Then L(3) u(xi) uh(xi) = a(G (x, xi) h, u uh

)but x = xi G(x, xi) =

ciNi(x) PWL basis functions on net {x0, . . . , xN} G vh Gh vh L(1)= u(xi)uh(xi) =0

38

• Notes:

1. True no matter how large or small N is.

2. Not true for F.D. equations

3. For this problem it is as if we knew the exact solution and fed this information in at thenodes. Thus the only error in the Galerkin approximation is the interpolation error:

||u uh|| h2

8||u|| = h

2

8||f || since u + f = 0.

4. It is not possible to use this technique for all problems as it relied on the fact thatG vh whichis not necessarily true for all problems. For further analysis see Strang and Fix, p. 3951.

||f || = maxi,j

supx(xi,xj)

|u(x)| <

Theorem: (Error in PWL approximation) Let f PC2,(0, 1) then ||f N

i=0fiNi(x)||

18h

2||f ||

Proof:

xi xi+1

fi+1f(x)

fiNi(x)

fi for any 0 i N

Let w(x) = (x xi)(x xi+1) = x2 (xi + xi+1)x+ xixi+1w = 2x (xi + xi+1) = 0 x =

(xi + xi+1

2

)w

(xi + xi+1

2

)=(xi+1 xi

2

)(xi xi+1

2

)

e(x) = f(x)N

i=0

Ni(x)fi

Claim: For each x [xi, xi+1] x [xi, xi+1] : e(x) = 12f (x)w(x)

Proof:1) x = xi , xi+1 any x suces.2) Choose an arbitrary x = x. Choose such that

(x) = e(x) w(x) = 0

39

• then (x) has 3 zeros on [xi, xi+1] andRolle x [xi, xi+1] : (x) = 0

x x

0

0 0 0

0 0 0

= 0

= 0

But

(x) = f (x) 2 = 0 = 1

2f (x)

maxx[xi,xi+1]

|e(x)| 12||f || max

x[xi,xi+1]|w(x)| 1

8h2||f ||

Theorem 2: Assume u c1[a, b]. Then there exists at least one point c (a, b) at which uh,x(c) =uexjx (c).

Proof:

MVT c (xi, xi+1) : u(xi+1) = u(xi) + (xi+1 xi)u,x(c)|| ||

uh(xi+1) uh(xi) + (xi+1 xi)u,x(c)

u,x(c) =uh(xi+1) uh(xi)

(xi+1 xi) = uh,x(x)

N i(x) = 1

xi+1 xi Ni+1(x) =

1xi+1 xi u

h,x(x) =

uh(xi+1) uh(xi)(xi+1 xi)

The Barlow points:If we dont know c then for linear elements uh,x at the midpoints are optimally accurate:

Assume u C3(a, b) and expand u about x = [xi, xi+1]:

u(xi+1) = u() + (xi+1 )u() + 12(xi+1 )2u() +

13!(xi+1 )3u(c1)

u(xi) = u() + (xi )u() + 12(xi )2u() +

13!(xi )3u(3)(c2)

Subtract and divide by xi+1 xi = hi+1u(xi+1) u(xi)

xi+1 xi = u() +

12(xi+1 + xi 2)(xi+1 xi)

xi+1 xi u() +O(h2i )

If = xi then uh,x(xi) =uh(xi+1) uh(xi)

xi+1 xi = u(xi) +

12hi+1u

(xi) +O(h2i )

If we choose =xi+1 + xi

2then uh,x(x) = u

(xi+1 + xi

2

)+ O(h2i )

Conclusion:For linear elements the midpoints are superconvergent with respect to derivatives, i.e. most accu-rate derivatives are calculated there. These are called the BARLOW points. (See Barlow, Int. J.

40

• of Num. Meth. Eng., p. 243251, 1976.

(B) Higher order elements:

Lagrange elements:We use the Lagrange polynomials to construct the basis functions:

Lda() =d+1b=1b=a

( b)/ d

b=1b =a

(a b)

Note: Lda(c) = ac just the right properties

d = 1: FIGURE

L11() =( 1)(1 1) =

12(1 ) = N11 ()

L12() =( (1))1 (1) =

12(1 + ) = N12 ()

d = 2: FIGURE

L21() =( 0)(1 0)

( 1)(1 1) =

12( 1) = N21 ()

L22() =( (1))( 1)(0 (1))(0 1) = (1

2) = N22 ()

L23() =( (1))

(1 (1))(1 0) =12(1 + ) = N23 ()

1 0 1

N23 ()

1 0 1

1 0 1

N21 ()

N22 ()

Note:3

k=1

N2k () =122 1

2 + 1 2 + 1

2 +

122 = 1

Can represent a constant function (or rigid body motion) exactly.

41

• d = 3 :

L31() =116

(1 )(92 1) = N31 ()

L32() =916

(3 1)(2 1) = N32 ()

L33() = 916

(3 + 1)(2 1) = N33 ()

L34() =116

(1 + )(92 1) = N34 ()

1 2 3 4

1 1/3 1/3 1

42

• Example:

u + f = 0u(0) = G u(1) = H

uh =N

i=0

uiNi(x)

u = NTu(W) a(uh, vh) = Hvh(b) + (f, vh)

xi1 1/2(xi + xi1) xi

xm

= 2hi (x xm)ddx =

2hi

keab =

xexe1

N a(x)Nb(x) dx

=

11

N a()2h N b()

2h h2d

ke11 =2h

11

{12(2 1)

}2d =

12h

11

42 4 + 1 d = 22h

[433 +

1

0

]=

73h

ke12 =2h

11

12(2 1)(2)d = 2

h

11

22 d = 2h 4

3

3

1

0

= 83h

ke13 =2h

11

12(2 1)1

2(2 + 1)d = +

24h

21

0

42 1d = 1h

[43

3

]10

=13h

ke22 =2h

11

(2)2d = 8h23

3

1

0

=163h

[ke] =13h

7 8 18 16 8

48 8 7

Let f = p : ue(x) = G + Hx + p(x x22

)

0 2

For 1 element h = 1

u0 u1 u2

13

16 88 7

u1

u2

=

83G + 23 p

H + p6 G3

f1 =

xexe1

p N1(x)dx = h2

11

(1 2)d = h[

3

3

]10

= 2h

3p

f2 = H +h

2

11

12( + 2)d = H + p

h

423

3

1

0

= H +h

6p

43

• u1

u2

=

7/48 1/6

1/6 1/3

8G + 2pG + 3H + 1/2p

=

7G/6G/6 + 7/24p + H/2 + 1/12p

4G/3 + 1/3pG/3 + H + p/16

=

G + 3/8p + H/2

G + 1/2p + H

N1(x) = 4x(1 x)

N2(x) = x(2x 1)

uh(x) = GN0(x) +(G +H/2 +

38p

)N1(x) + (G + p/2 + H)N2(x)

= G + (H/2N1 (x) +HN2 (x)) +(38pN1 (x) +

p

2N2 (x)

)

= G +H(12(4x 4x2)+ 2x2 x)+ p(3

8(4x 4x2)+ 1

2(2x2 x)

)

= G +Hx+p

2{3x 3x2 + 2x2 x}

= G +Hx+p

2{2x x2}

= G +Hx+ p(x x

2

2

)the exact solution

44